Perfect Numbers - A Case Study

Perfect numbers are those numbers that equal the sum of all their divisors including 1 and excluding the number itself.

Most numbers do not fit this description. At the heart of every perfect number is a Mersenne prime. All of the other divisors are either powers of 2 or powers of 2 times the Mersenne prime.

Let's examine the number 496 - one of the known perfect numbers. In order to demonstrate that 496 is a perfect number, we must show that

496 = (the sum of all its divisors including 1 and excluding 496)

We might just start by dividing and working out the divisors the long way. Or, we might begin by noting that, in the notation that includes a Mersenne prime,

496 = 24 (2⁵ - 1) = 2⁴ x 31.

From this expression, we may easily obtain the divisors of 496, namely:

1, 2, 2², 2³, 2⁴, 31, (2×31), (2²×31), (2³×31), and (2⁴×31).

The sum of all the divisors excluding 496 is then

1 + 2 + 2² + 2³ + 2⁴ + 31× (1 + 2 + 2²+ 2³)

In order to break this mess down a bit, let us examine the partial sum

u = 1 + 2 + 2² + 2³ + 2⁴

If we multiply by 2 (i.e., 2u = 2 + 2² + 2³ + 2⁴ + 2⁵) then subtract, we find

u = 2u - u = 2⁵ - 1 = 31

Thus, the sum of the divisors becomes

= 31 + 31(2⁴ - 1)
= 2⁴×31
= 496

as we were to show.
To show that any perfect number may be broken down in this way, we let

n_P = 2^c(2^c+1 - 1)

be a perfect number with (2c+1 -1) being the embedded Mersenne prime. Then the divisors of n_P are

1, 2, 2², ... , 2^c, 2^c+1 - 1, 2(2^c+1 - 1), ..., 2^c(2^c+1 - 1)

The sum of all the divisors excluding np is then

1 + 2 + 2² + ... + 2^c + (2^c+1 - 1) (1 + 2 + ... + 2^c-1)

Again, we examine the partial sum:

u = 1 + 2 + 2² + ... + 2^c.

We multiply by 2 (i.e., 2u = 2 + 2² + ... + 2^c + 2^c+1) then subtract:

u = 2u - u = 2^c+1 - 1

The sum of the divisors becomes:

= (2^c+1 - 1) + (2^c+1 - 1)(2^c - 1)
= 2^c(2^c+1 - 1)

as we were to show.